Answer
$$ y=e(x+2).$$
Work Step by Step
We have
$$ y'=e^{x+2}\Longrightarrow y'(-1)=e.$$
Then the slope of the tangent line at $ x=-1$ is $ m= e $. Hence the equation of the tangent line is given by
$$ y=mx+c=ex+c.$$
Since $ y(-1)=e $, then $ c=2e $, hence the tangent line at $ x=-1$ is given by
$$ y=ex+2e=e(x+2).$$